Chapter 8 - Section 8.2 - Inconsistent and Dependent Systems and Their Applications - Exercise Set - Page 903: 25

a) The system represented by A is $4w-2x+2y-3z=0$, $7w-x-y-3z=0$, and $ w+x+y-z=0$. b) The system’s solution set is $\left( 0.5z,0,0.5z,z \right)$.

(a) Consider the augmented matrix: $\left[ \begin{matrix} 4 & -2 & 2 & -3 & 0 \\ 7 & -1 & -1 & -3 & 0 \\ 1 & 1 & 1 & -1 & 0 \\ \end{matrix} \right]$. Use back-substitution to express $ w,x,y,z $. $\begin{align} & 4w-2x+2y-3z=0 \\ & 7w-x-y-3z=0 \\ & w+x+y-z=0 \end{align}$ Therefore, the system represented by A is $4w-2x+2y-3z=0$, $7w-x-y-3z=0$, and $ w+x+y-z=0$. (b) Consider the following augmented matrix: $\left[ \begin{matrix} 4 & -2 & 2 & -3 & 0 \\ 7 & -1 & -1 & -3 & 0 \\ 1 & 1 & 1 & -1 & 0 \\ \end{matrix} \right]$. Now, use row-echelon method by applying row operation, Interchange ${{R}_{1}}\leftrightarrow {{R}_{3}}$ $\left[ \begin{matrix} 1 & 1 & 1 & -1 & 0 \\ 7 & -1 & -1 & -3 & 0 \\ 4 & -2 & 2 & -3 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to {{R}_{2}}-7{{R}_{1}}$, ${{R}_{3}}\to {{R}_{3}}-4{{R}_{1}}$, $\left[ \begin{matrix} 1 & 1 & 1 & -1 & 0 \\ 0 & -8 & -8 & 4 & 0 \\ 0 & -6 & -2 & 1 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to -\frac{1}{4}{{R}_{2}}$, $\left[ \begin{matrix} 1 & 1 & 1 & -1 & 0 \\ 0 & 2 & 2 & -1 & 0 \\ 0 & -6 & -2 & 1 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}+3{{R}_{2}}$, $\left[ \begin{matrix} 1 & 1 & 1 & -1 & 0 \\ 0 & 2 & 2 & -1 & 0 \\ 0 & 0 & 4 & -2 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to \frac{1}{2}{{R}_{2}}$, ${{R}_{3}}\to \frac{1}{4}{{R}_{3}}$, $\left[ \begin{matrix} 1 & 1 & 1 & -1 & 0 \\ 0 & 1 & 1 & -0.5 & 0 \\ 0 & 0 & 1 & -0.5 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$, $\left[ \begin{matrix} 1 & 0 & 0 & -0.5 & 0 \\ 0 & 1 & 1 & -0.5 & 0 \\ 0 & 0 & 1 & -0.5 & 0 \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$, $\left[ \begin{matrix} 1 & 0 & 0 & -0.5 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & -0.5 & 0 \\ \end{matrix} \right]$ Now, use back-substitution to express $ w,x,y,z $: $\begin{align} & w-0.5z=0 \\ & x=0 \\ & y-0.5z=0 \end{align}$ Now, in the above linear system, From the third equation we get, $\begin{align} & y-0.5z=0 \\ & y=0.5z \end{align}$ From second equation we get, $ x=0$ From first equation we get, $\begin{align} & w-0.5z=0 \\ & w=0.5z \end{align}$ In the provided matrix, there are four variables and three equations, so substitute the value of $ z=t $. Therefore, the system’s solution set is $\left( 0.5z,0,0.5z,z \right)$